A Polynomial Time Algorithm to Find Rational General Solutions of First Order Autonomous ODEs

Transcription

1 J. Symbolic Computation (2005) 11, 1 22 A Polynomial Time Algorithm to Find Rational General Solutions of First Order Autonomous ODEs Ruyong Feng and Xiao-Shan Gao Key Lab of Mathematics Mechanization, Institute of Systems Science, AMSS Academia Sinica, Beijing , China (Received October, 2004) We give a necessary and sufficient condition for an algebraic ODE to have a rational type general solution. For a first order autonomous ODE F = 0, we give an exact degree bound for its rational solutions, based on the connection between rational solutions of F = 0 and rational parameterizations of the plane algebraic curve defined by F = 0. For a first order autonomous ODE, we further give a polynomial time algorithm to compute a rational general solution if it exists based on the computation of Laurent series solutions and Padé approximants. Experimental results show that the algorithm is quite efficient. Keywords. Rational general solution, first order autonomous ODE, rational parameterizations, Laurent series, Padé approximants, polynomial time algorithm. 1. Introduction In the pioneering papers (Risch, 1969; Risch, 1970), Risch described a method to find the elementary integral of udx where u is an elementary function. In (Kovacic, 1986), Kovacic presented an effective method to find Liouvillian solutions for second order linear homogeneous differential equations and Riccati equations. In (Singer, 1981), Singer established the general framework for finding Liouvillian solutions for general linear homogeneous ODEs. Many other interesting results on finding Liouvillian solutions of linear ODEs were reported in (Abramov and Kvashenko, 1991; Bronstein and Lafaille, 2002; Cormier, 2001; Ulmer and Calmet, 1990; van Hoeij et al., 1999; Van der Put and Singer, 2003). In (Li and Schwarz, 2001), Li and Schwarz gave the first method to find rational solutions for a class of partial differential equations. Most of these results are limited to the linear case or some special type nonlinear equations. There seems exist no general methods to find closed form solutions for nonlinear differential equations. With respect to the particular ODEs of the form y = R(x, y) where R(x, y) is a rational function, Darboux and Poincaré made important contributions (Poincaré, 1897). More recently, Cerveau, Neto and Carnicer also made important progresses (Cerveau and Lins Neto, 1991; Carnicer, 1994). In particular, Carnicer gave the degree bound of algebraic solutions in the nondicritical case. In (Cano, 2003), Cano proposed an algorithm to find their polynomial solutions. In (Singer, 1992), Singer studied the Liouvillian first integrals of differential equations. In (Bronstein, 1992), Bronstein /90/ $03.00/0 c 2005 Academic Press Limited

2 2 Ruyong Feng and Xiao-Shan Gao gave an effective method to compute rational solutions of the Ricatti equations. For a general first order differential equation, Eremenko proved that there exists a degree bound of rational solutions in (Eremenko, 1998), but the proof is not constructive. In (Hubert, 1996), Hubert gave a method to compute a basis of the general solutions of first order ODEs and applied it to study the local behavior of the solutions. In this paper, we try to find rational type general solutions to algebraic ODEs. For example, the general solution for dy dx + y2 = 0 is y = 1 x+c, where c is an arbitrary constant. The motivation of finding the rational general solutions to algebraic ODEs is as follows. Converting between implicit representation and parametric representation of (differential) varieties is one of the basic topics in (differential) algebraic geometry. In the differential case, implicitization algorithms were given in (Gao, 2003). As far as we know, there exist no general results on parametrization of differential varieties. The results in this paper could be considered as a first step to the rational parametrization problem for differential varieties. Three main results are given in this paper. In Section 2, for non-negative integers n and m, we define a differential polynomial D n,m in variable y such that the solutions of the ODE D n,m = 0 are rational solutions whose numerator and denominator are of degrees less than n and m respectively. Based on this, we give a sufficient and necessary condition for an algebraic ODE to have a rational general solution. By treating the variable and its derivative as independent variables, a first order autonomous (constant coefficients) ODE defines an algebraic plane curve. In Section 3, we show that a nontrivial rational solution of a first order autonomous ODE and its derivative provides a proper parametrization of its corresponding algebraic curve. From this result, we may obtain an exact degree bound for its rational solutions. In Section 4, we give a detailed analysis of the structural properties of a first order autonomous ODE with a rational solution. These properties give necessary conditions for a first order autonomous ODE to have rational solutions. We also present a polynomial-time algorithm to compute the first 2n + 1 terms of a Laurent series solution to a first order autonomous ODE in certain case. These results and Padé approximants are finally used to give a polynomial time algorithm to find a rational general solution for a first order autonomous ODE. For the first order autonomous ODE, finding the solutions is equivalent to finding the integration of algebraic function, because F ( dy dx, y) = 0 implying that there exists a G(z 1, z 2 ) such that G( dx dy, y) = 0. In (Davenport, 1981; Trager, 1984), Davenport and Trager gave an algorithm to find the integration of algebraic functions. In (Bronstein, 1990), Bronstein generalized Trager s algorithm to elementary functions. Their algorithms can compute elementary integration but the complexity in the worst-case is exponential. Here, our algorithm is equivalent to an algorithm for finding a special algebraic integration for algebraic functions and the complexity is polynomial. The algorithm is implemented in Maple and experimental results show that the algorithm is very efficient. For two hundred randomly generated first order autonomous ODEs, the algorithm can immediately (without computation) decide that the ODEs do not have rational general solutions using the necessary conditions presented in Section 4. For large first order autonomous ODEs with rational general solutions, the algorithm can find their rational solutions efficiently. This paper is an essential improvements of our ISSAC 04 paper (Feng and Gao, 2004). The main improvement is in Section 4, where we present a polynomial time algorithm

3 Rational Solutions of ODEs 3 to find rational general solutions to first order autonomous ODEs by proving structural properties of the ODEs and a new algorithm to compute the Laurent series solutions. While, the algorithm in (Feng and Gao, 2004) is exponential. Experimentally, the average running time of our new algorithm for the same set of ODEs with rational solutions is about thirty times faster and the new algorithm can solve much larger problems. Due to the structural properties, the new algorithm gives immediate negative answer to almost all randomly generated ODEs. Another advantage of the new algorithm is that we do not need to work over the fields of algebraic numbers. In Sections 2 and 3, Lemmas 2.9 and 3.2, Theorems 2.5, 2.10 and 3.8 are also new results. 2. Rational general solutions of algebraic ODEs 2.1. Definition of rational general solutions In the following, let K = Q(x) be the differential field of rational functions in x with differential operator d dx and y an indeterminate over K. Let Q be the algebraic closure of the rational number field Q. We denote by y i the i-th derivative of y. We use K{y} to denote the ring of differential polynomials over the differential field K, which consists of the polynomials in the y i with coefficients in K. All differential polynomials in this paper are in K{y}. Let Σ be a system of differential polynomials in K{y}. A zero of Σ is an element in a universal extension field of K, which vanishes every differential polynomial in Σ (Ritt, 1950). In this paper, we also assume that the universal extension field of K contains an infinite number of arbitrary constants. The totality of the zeros in K is denoted by Zero(Σ). Let P K{y}/K. We denote by ord(p ) the highest derivative of y in P, called the order of P. Let o = ord(p ) > 0. We may write P as follows P = a d y d o + a d 1 y d 1 o a 0 where a i are polynomials in y, y 1,..., y o 1 and a d 0. a d is called the initial of P and S = P y o is called the separant of P. The k-th derivative of P is denoted by P (k). Let S be the separant of P, o = ord(p ) and an integer k > 0. Then we have P (k) = Sy o+k R k (2.1) where R k is of lower order than o + k. Let P be a differential polynomial of order o. A differential polynomial Q is said to be reduced with respect to P if ord(q) < o or ord(q) = o and deg(q, y o ) < deg(p, y o ). For two differential polynomials P and Q, let R = prem(p, Q) be the differential pseudoremainder of P with respect to Q. We have the following differential remainder formula for R (Kolchin, 1973; Ritt, 1950) JP = i B i Q (i) + R where J is a product of certain powers of the initial and separant of Q and B i, R are differential polynomials. Moreover, R is reduced with respect to Q. For a differential polynomial P with order o, we say that P is irreducible if P is irreducible when P is treated as a polynomial in K[y, y 1,..., y o ]. In this paper, when we say a differential polynomial irreducible, we always mean that it is irreducible over Q(x)[y, y 1,..., y o ].

4 4 Ruyong Feng and Xiao-Shan Gao Let P K{y}/K be an irreducible differential polynomial and Σ P = {A K{y} SA 0 mod {P }}. (2.2) where{p } is the perfect differential ideal generated by P (Kolchin, 1973; Ritt, 1950). Ritt proved that (Ritt, 1950) Lemma 2.1. Σ P is a prime differential ideal and a differential polynomial Q belongs to Σ P iff prem(q, P ) = 0. Let Σ be a non-trivial prime ideal in K{y}. A zero η of Σ is called a generic zero of Σ if for any differential polynomial P, P (η) = 0 implies that P Σ. It is well known that an ideal Σ is prime iff it has a generic zero (Ritt, 1950). When we say a constant, we mean it is in the constant field of the universal extension field of K. The following definition of general solution is due to Ritt. Definition 2.2. Let F K{y}/K be an irreducible differential polynomial. A general solution of F = 0 is defined as a generic zero of Σ F. A rational general solution of F = 0 is defined as a general solution of the form where a i, b j F = 0. ŷ = a nx n + a n 1 x n a 0 x m + b m 1 x m b 0 (2.3) are constants. When m = 0, ŷ is called a polynomial general solution of Notation 2.3. deg x (ŷ):=max{n, m} where ŷ is as in (2.3) and a n 0. As a consequence of Lemma 2.1, we have Lemma 2.4. Let F K{y}/K be an irreducible differential polynomial with a generic solution η. Then for a differential polynomial P we have P (η) = 0 iff prem(p, F ) = 0. A general solution of F = 0 is usually defined as a family of solutions with o independent parameters in a loose sense where o = ord(f ). The definition given by Ritt is more precise. Theorem 6 in section 12, chapter 2 in (Kolchin, 1973) tells us that Ritt s definition of general solution is equivalent to the definition in the classical literature. The universal constant extension of Q is obtained by first adding an infinite number of arbitrary constants to Q and then taking the algebraic closure A Criterion for existence of rational general solutions For non-negative integers n and m, let D n,m be the following differential polynomial in y: ( n+1 ) ( n+1 ) ( 0 yn+1 1 yn... n+1 ) m yn+1 m (n+2 ) ( n+2 ) ( 0 yn+2 1 yn+1... n+2 ) m yn+2 m ( n+m+1 ) ( n+m+1 ) ( 0 yn+m+1 1 yn+m... n+m+1 ) m yn+1

5 Rational Solutions of ODEs 5 where ( n k) are binomial coefficients and ( n k) = 0 for k > n. First, we can prove that Dn,m is irreducible. Theorem 2.5. D n,m is an irreducible differential polynomial. Proof. Since D n,m is a homogeneous polynomial with degree m+1 and includes a term yn+1 m+1, we need only to prove the result in the case y n+2 = y n+3 = = y n+m = 0. We will use D n,m to denote the polynomial obtained by replacing y i (n + 1 < i < n m) with 0 in D n,m. By the computation process, we have D n,m = yn+1 m+1 + ( 1)m y n+1+m D where D is a polynomial including a term n m yn m and with total degree not greater than m. Because y n+1+m is linear in D n,m, by Eisenstein s Criterion (Van der Waerden, 1970), D n,m is irreducible. Note that when m = 0, D n,0 = y n+1, whose solutions are c n x n + c n 1 x n c 0 where c i are constants. In the general case, we have Lemma 2.6. The solutions of D n,m = 0 have the following form: where a i, b j are constants. Proof. Let Then we have B = ŷ = a nx n + a n 1 x n a 0 b m x m + b m 1 x m b 0 D n,m B = x m x m mx m 1 (m 1)x m m!x (m 1)!... 0 m! (x m y) (n+1) (x m 1 y) (n+1)... y (n+1) (x m y) (n+2) (x m 1 y) (n+2)... y (n+2) (x m y) (2n) (x m 1 y) (2n)... y (2n) which is a Wronskian determinant for (x m y) (n+1), (x m 1 y) (n+1), y (n+1) (Ritt, 1950). Hence we have D n,m (ŷ) B = 0 if and only if there exist constants b m, b m 1,, b 0, not all of them equal to 0, such that b m (x m ŷ) (n+1) + b m 1 (x m 1 ŷ) (n+1) + + b 0 = 0. Since B = 0, D n,m (ŷ) B = 0 D n,m (ŷ) = 0. As a consequence, D n,m (ŷ) = 0 ((b m x m + b m 1 x m b 0 )ŷ) (n+1) = 0 ŷ = a nx n + + a 0 b m x m + + b 0 where a i are constants. By Lemma 2.6, we can prove the following theorem easily.

6 6 Ruyong Feng and Xiao-Shan Gao Theorem 2.7. Let F be an irreducible differential polynomial. Then the differential equation F = 0 has a rational general solution ŷ iff there exist non-negative integers n and m such that prem(d n,m, F ) = 0. Proof. ( ) Let ŷ = P (x) Q(x) be a rational general solution of F = 0. Let n deg(p (x)) and m deg(q(x)). Then from Lemmas 2.4 and 2.6 D n,m (ŷ) = 0 D n,m Σ F prem(d n,m, F ) = 0. ( ) By Lemma 2.1, prem(d n,m, F ) = 0 implies that D n,m Σ F. Assume that m is the least integer such that D n,m Σ F. Then all the zeros of Σ F must have the form ȳ = ānx n + ā n 1 x n ā 0 bm x m + b m 1 x m b 0. In particularly, the generic zero of Σ F has the following form ŷ = a nx n + a n 1 x n a 0 b m x m + b m 1 x m b 0. Moreover, b m 0. Otherwise, we would have D n,m 1 (ŷ) = 0 which implies that D n,m 1 Σ F, a contradiction. So the generic zero has the form (2.3). In the above theorem, let m = 0. Then we have the following corollary: Corollary 2.8. Let F be an irreducible differential polynomial. Then the differential equation F = 0 has a polynomial general solution ŷ iff there exists a non-negative integer n such that prem(y n, F ) = 0. Given a differential equation F = 0, if we know the degree bound N of the rational general solution of it, then we can decide whether it has a rational general solution or not by computing prem(d n,m, F ) for n, m = 1,, N. However, for higher order ODEs or ODEs with variate coefficients, we do not know this degree bound. In the following, we will show that a special case can be solved elegantly. In the Section 3, for the first order autonomous ODEs, we will give an exact degree bound for its rational solutions. Lemma 2.9. Let y = a n x n + a n 1 x n a 0 where a i are arbitrary constants. Let y i be the i-th derivative of y with respect to x and y 0 = y. Then for i = 0,, n, we have x n i+1 a i = ( 1) n i i!(n i)! (y i x )n i. Proof. We note that ( yi x )(n i) = ( i!ai x )(n i). Then by the computation directly, we can prove the lemma. Theorem Let F be an irreducible differential polynomial and n = ord(f ). Then F = 0 has a polynomial general solution with degree n iff F can be rewritten as the following form: F = p(x)( c i0,i 1,,i n P i0 0 P i1 1 P in n ) (2.4) n i xn i+1 where p(x) K, P i = ( 1) i!(n i)! ( yi x )(n i) and c i0,i 1,,i n Q

7 Rational Solutions of ODEs 7 Proof. ( ) Suppose that ŷ = â n x n + â n 1 x n â 0 is a polynomial general solution of F = 0 with degree n. Then (â 0, â 1,, â n ) satisfies an algebraic equation: G(â 0, â 1,, â n ) = 0 where G Q[z 0,, z n ] and is irreducible. Let G = G(P 0, P 1,, P n ) n i xn i+1 where P i = ( 1) i!(n i)! ( yi x )(n i). It is easy to know that P i Q[x, y 0, y 1,, y n ], y k (k i) appears linearly in P i, and the coefficient of y i is a nonzero rational number. Hence G Q[x, y 0,, y n ] is an irreducible polynomial. By Lemma 2.9, we have that â i = P i (ŷ) for i = 0,, n. So if we regard G as a differential polynomial, we have that G(ŷ) = G(â 0, â 1,, â n ) = 0. Because F and G are all irreducible and with order n, F = p(x) G where p(x) K. ( ) Suppose that F has the form (2.4). Let G = c i0,i 1,,i n z i0 0 zi1 1 zin n Q[z 0,, z n ], then p(x)g(p 0, P 1,, P n ) = F. Let ŷ = â n x n +â n 1 x n 1 + +â 0 where (â 0, â 1,, â n ) is a generic zero of G = 0 and â n 0, â i are arbitrary constants. We will prove that ŷ is a polynomial general solution. By Lemma 2.9, we know that F (ŷ) = p(x)g(p 0 (ŷ),, P n (ŷ)) = p(x)g(â 0,, â n ) = 0. Assume that H K{y} satisfies H(ŷ) = 0. Let R = prem(h, F ). Then R(ŷ) = 0. Assume that R 0, we will get a contradiction. Since y k (k i) appear linearly in P i and the coefficient of y i is a nonzero rational number, we can rewrite R as the form R = b j0,j 1,,j n P j0 0 P j1 1 P n jn + h(x) where b j0,j 1,,j n, h(x) K. Let R = b j0,j 1,,j n z j0 0 zj1 1 zjn n +h(x). Then by Lemma 2.9, R(â 0,, â n ) = R(ŷ) = 0. Hence R = M G where M K[z 0,, z n ] because (â 0,, â n ) is a generic zero of G = 0. That is, p(x)r = M(P 0,, P n ) F, this is impossible. Hence R 0 which implies that H Σ F where Σ F as in (2.2). So ŷ is a generic zero of Σ F. From the definition of the general solution, ŷ is a polynomial general solution. 3. Rational general solution of first order autonomous ODE In the following sections, F will always be a non-zero first order autonomous differential polynomial with coefficients in Q and irreducible in the polynomial ring Q[y, y 1 ]. We call a rational solution ȳ of F = 0 nontrivial if deg x (ȳ) > 0. It is a trivial fact that for an autonomous ODE, the solution set is invariant by a translation of the independent variable x. Moreover, we have the following fact. Lemma 3.1. Let ȳ = ānxn +...+ā 0 x m b 0 be a nontrivial solution of F = 0, where ā i, b j are constants and ā n 0. Then ŷ = ān(x + c) n ā 0 (x + c) m b 0 is a rational general solution of F = 0, where c is an arbitrary constant. Proof. It is easy to show that ŷ is still a zero of Σ F. For any G K{y} satisfying G(ŷ) = 0, let R = prem(g, F ). Then R(ŷ) = 0. Suppose that R 0. Since F is irreducible and deg(r, y 1 ) < deg(f, y 1 ), there are two differential polynomials P, Q K{y} such that

8 8 Ruyong Feng and Xiao-Shan Gao P F + QR K[y] and P F + QR 0. Thus (P F + QR)(ŷ) = 0. Because c is an arbitrary constant which is transcendental over K, we have P F + QR = 0, a contradiction. Hence R = 0 which means that G Σ F. So ŷ is a generic zero of Σ F. Note that when we say constant, we mean a constant in the constant field of universal extension field of K. We do not know whether its constant field is exactly the universal constant extension of Q. The following lemma shows that for fist order autonomous ODE, the constants in its rational general solution can be chosen in the universal constant extension of Q.. Lemma 3.2. The constants in a rational general solution of a first order autonomous algebraic ODE can be chosen in the universal constant extension of Q. Proof. From Lemma 3.1, we need only to prove ā i, b j can be chosen in Q. Substituting an arbitrary rational function (2.3) into F = 0, we have F = P (x)/q(x), where P (x) and Q(x) are polynomials in x whose coefficients are polynomials in a i, b j. Not that Q(x) do not always vanish because the rational function is of the form (2.3). Let P S be the coefficients of P (x). Then (ā 0,, ā n, b 0,, b m 1 ) must be a zero of P S with a n 0. Since Q is algebraical closure, we can get a zero of P S in Q with a n 0. Hence ā i, b j can be chosen in Q. Lemma 3.1 reduces the problem of finding a rational general solution to the problem of finding a nontrivial rational solution. In what below, we will show how to find a nontrivial rational solution. First of all, we decide the degree of a nontrivial rational solution Parametrization of algebraic curves In this subsection, we will introduce some basic concepts on the parametrization of an algebraic plane curve. Let F (x, y) be a polynomial in Q[x, y] and irreducible over Q. Definition 3.3. (r(t), s(t)) is called a parametrization of F (x, y) = 0 if F (r(t), s(t)) 0 where r(t), s(t) Q(t) and not all of them are in Q. A parametrization (r(t), s(t)) is called proper if Q(r(t), s(t)) = Q(t). Lüroth s Theorem guarantees that we can always obtain a proper parametrization from an arbitrary rational parametrization (Van der Waerden, 1970; Gao and Chou, 1991). Lemma 3.4. A proper parametrization has the following properties (Sendra and Winkler, 2001): 1 deg t (r(t)) = deg(f, y) 2 deg t (s(t)) = deg(f, x) 3 If (p(t), q(t)) is another proper parametrization of F (x, y), then there exists f(t) = at+b such that p(t) = r(f(t)), q(t) = s(f(t)) where a, d, b, c are elements in Q ct+d satisfying ad bc. 4 Assume that r(t) = r1(t) r 2(t) s1(t) and s(t) = s where r 2(t) i(t), s i (t) Q[t]. Let R(x, y) be the Sylvester-resultant of r 2 (t)x r 1 (t) and s 2 (t)y s 1 (t) with respect to t. Then R(x, y) = λf where λ Q.

9 Rational Solutions of ODEs Degree for rational solutions of first order autonomous ODEs Since F has order one and constant coefficients, we can regard it as an algebraic polynomial in y, y 1. Notation 3.5. We use F (y, y 1 ) to denote F as an algebraic polynomial in y and y 1 which defines an algebraic curve. If ȳ = r(x) is a nontrivial rational solution of F = 0, then (r(x), r (x)) can be regarded as a parametrization of F (y, y 1 ) = 0. Moreover, we will show that (r(x), r (x)) is a proper parametrization of F (y, y 1 ) = 0. Lemma 3.6. Let f(x) = p(x) q(x) Q be a rational function in x such that gcd(p(x), q(x)) = 1. Then Q(f(x)) Q(f (x)). Proof. If f (x) Q then the result is clearly true. Otherwise, f(x), f (x) are transcendental over Q. If Q(f(x)) = Q(f (x)), from the Theorem in Section 63 of (Van der Waerden, 1970), we have f(x) = af (x) + b cf (x) + d where a, b, c, d Q. Then p(x) q(x) = a(p (x)q(x) p(x)q (x)) + bq(x) 2 c(p (x)q(x) p(x)q (x)) + dq(x) 2 which implies that q(x) cp(x)q (x) because gcd(p(x), q(x)) = 1. So c = 0 or q (x) = 0 which implies that f(x) = ( a d )f (x) + b d or p(x) = c1p (x)+c 2 c 3p (x)+c 4 where c i Q. This is impossible, because f(x) is a rational function and p(x) is a nonconstant polynomial if q(x) Q. Theorem 3.7. Let f(x) be the same as in Lemma 3.6. Then Q(f(x), f (x)) = Q(x). Proof. From Lüroth s Theorem (Van der Waerden, 1970), there exists g(x) = u(x) v(x) such that Q(f(x), f (x)) = Q(g(x)), where u(x), v(x) Q[x], gcd(u(x), v(x)) = 1. We may assume that deg(u) > deg(v). Otherwise, we have u v = c + w v where c Q and deg(w) < deg(v), and v w is also a generator of Q(g(x)). Then we have f(x) = p 1(g(x)) q 1 (g(x)), f (x) = p 2(g(x)) q 2 (g(x)) = g (x)(p 1q 1 p 1 q 1) q 2 1 which implies that g (x) Q(g(x)). If g (x) / Q, we have [ Q(x) : Q(g (x))] = [ Q(x) : Q(g(x))][ Q(g(x)) : Q(g (x))]. However, we have [ Q(x) : Q(g(x))] = deg(u) and [ Q(x) : Q(g (x))] 2 deg(u) 1. Hence [ Q(x) : Q(g(x))] = [ Q(x) : Q(g (x))]. That is, Q(g (x)) = Q(g(x)), a contradiction by Lemma 3.6. Hence, g (x) Q which implies that g(x) = ax + b. The above theorem implies that (ȳ, ȳ 1 ) is a proper parametrization of F (y, y 1 ) = 0 if ȳ is a nontrivial rational solution of F = 0. By Lemma 3.4, we have

10 10 Ruyong Feng and Xiao-Shan Gao Theorem 3.8. Assume that ȳ = P (x) Q(x) is a nontrivial rational solution of F = 0. Let U = P (x) yq(x), V = P (x) yq (x) y 1 Q(x) and R be the Sylvester-resultant of U and V with respect to x. Then R = λf where λ is a non-zero element in Q. The above theorem proves the intuition that if y = f(x) is a rational solution of F = 0 then F may be obtained by eliminating x from y = f(x) and y 1 = f (x). It is easy to show that this result is not valid anymore for algebraic solutions. Lemma 3.9. Let f(x) be the same as in Lemma 3.6. Then deg x (f(x)) 1 deg x (f (x)) 2 deg x (f(x)). Proof. Since f(x) is rational, it is clear that deg x (f (x)) 2 deg x (f(x)). If q(x) Q, then deg x (( p(x) q(x) ) ) = deg x ( p(x) q(x) ) 1. Assume that q(x) / Q. Let q(x) = (x a 1 ) α1 (x a 2 ) α2... (x a r ) αr. Then ( p(x) r q(x) ) = p (x ai ) p( i=1 j i α i(x a j )) (x a 1 ) α1+1 (x a 2 ) α (x a r ) αr+1 Since p (x a i ) p( r i=1 j i α i(x a j )) and (x a 1 ) α1+1 (x a 2 ) α (x a r ) αr+1 have no common divisors, we have deg x (( p(x) q(x) ) ) = max{deg(p) + r 1, deg(q) + r} > deg x ( p(x) q(x) ) 1. From Lemmas 3.4 and 3.9, we have proved the following key theorem. Theorem If F = 0 has a rational general solution ŷ, then we have { deg x (ŷ) = deg(f, y 1 ) deg(f, y 1 ) 1 deg(f, y) 2 deg(f, y 1 ) From Theorems 2.7 and 3.10, we have the following corollary: Corollary Let d = deg(f, y 1 ). Then F = 0 has a rational general solution iff prem(d d,d, F ) = 0. Remark In chapter X, vol 2(Forsyth, 1959), there is a necessary and sufficient condition for a first order autonomous ODE to have a uniform solution by analytical consideration. However, the condition given here is simpler. Remark We may find a rational solution to F = 0 as follows. Let d = deg(f, y 1 ). As in the Lemma 3.2, substituting an arbitrary rational function (2.3) of degree d into F = 0, we have F = P (x)/q(x), where P (x) and Q(x) are polynomials in x whose coefficients are polynomials in a i, b j. Note that Q(x) does not always vanish because the rational function is of the form (2.3). Let P S be the coefficients of P (x) as a polynomial in x. Then (2.3) is a rational solution to F = 0 iff a i, b j are zeros of the polynomial equations in P S. This method is not efficient for large d since it involves the solution of a nonlinear algebraic equation system in 2d variables. We will give more efficient algorithms in Sections 4.4.

11 Rational Solutions of ODEs 11 In the paper (Sendra and Winkler, 1997), Sendra and Winkler proved that for a rational algebraic curve defined by a polynomial over Q which is irreducible over Q, it can be parameterized over an extension field of Q with degree at most two. Theorem 3.14 will tell us that F (y, y 1 ) = 0 can always be parameterized over Q. Theorem If F = 0 has a nontrivial rational solution, then the coefficients of the nontrivial rational solution can be chosen in Q. Proof. From Theorem 3.1 in (Sendra and Winkler, 1997) and Theorem??, we know that there exists a nontrivial rational solution r(x) of F = 0 whose coefficients belong to Q(α) where α 2 Q. From Lemma 3.1, we can assume that r(x) = αp1(x)+p2(x) x m +αq where 1(x)+q 2(x) p i (x), q j (x) Q(x). Assume that αp 1 (x)+p 2 (x) and x m +αq 1 (x)+q 2 (x) have no common divisors over Q(α)[x]. We further assume that deg(q j (x)) m 2, which may be achieved by a proper linear transformation. It is easy to check that r(x) = αp1(x)+p2(x) x m αq 1(x)+q 2(x) is also a nontrivial rational solution of F = 0. Since both (r(x), r (x)) and ( r(x), r (x)) are proper parametrizations of F (y, y 1 ) = 0, there exists an f(x) such that r(x) = r(f(x)) and r (x) = r (f(x)). Since r (x) = f (x) r (f(x)), we have f (x) = 1 which implies that f(x) = x + c where c Q(α). Thus αp 1 (x) + p 2 (x) x m + αq 1 (x) + q 2 (x) = αp 1 (x + c) + p 2 (x + c) (x + c) m αq 1 (x + c) + q 2 (x + c). Since αp 1 (x) + p 2 (x) and x m + αq 1 (x) + q 2 (x) have no common divisors, we have x m + αq 1 (x) + q 2 (x) = (x + c) m αq 1 (x + c) + q 2 (x + c). If m > 0, we have c = 0 because deg(q j (x)) m 2, which implies that p 1 (x) = q 1 (x) = 0. If m = 0, then r(x) is a polynomial. We can assume that r(x) = (a n α+ã n )x n +αp 1 (x)+ p 2 (x) where p i (x) Q(x), deg(p i (x)) n 2 and a n, ã n Q, where at least one of a n and ã n is not 0. In a similar way, we have a n = 0 and p 1 (x) = A polynomial-time algorithm for first order autonomous ODEs In this section, we will give an effective method with polynomial complexity Structure of first order autonomous ODEs with rational solutions In order to present the algorithm, we need to analysis the structure of a first order autonomous ODE with nontrivial rational solutions. In this section, F = 0 is always a first order autonomous ODE. We write F as the following forms: F = A d (y)y1 d + A d 1 (y)y1 d A 0 (y) (4.1) F = F d(y, y 1 ) + F d 1 (y, y 1 ) + + F d (y, y 1 ) (4.2) where A i (y) are polynomials in y, F i (y, y 1 ) is the homogeneous part of F with total degree i, and d d. It is clear that d = tdeg(f ) is the total degree of F. We are going to assume that F = 0 has a nontrivial rational solution of the form ȳ = P (x), where n = deg(p (x)), m = deg(q(x)). (4.3) Q(x)

12 12 Ruyong Feng and Xiao-Shan Gao By Theorem 3.10, we have d = max{n, m}. As a corollary of the theorem on page 311 of (Forsyth, 1959), we have Lemma 4.1. If F = 0 of the form (4.1) has a nontrivial rational solution, then deg(a i (y)) 2(d i) for i = 0,, d. In fact, the above lemma is still true for a first order ODE with variate coefficients which has no movable singularities (Matsuda, 1980). Theorem 4.2. Assume that F = 0 of the form (4.2) has a nontrivial rational solution of form (4.3). We have (1) If n > m, then d = deg(f, y) + 1. (2) If n m, then d = deg(f, y). Proof. In this proof, we use F (y, y 1 ) to denote F. Replacing y 1 by ty + z, F d(y, ty + z) = d C d,j y d j (ty + z) j = ( d j=0 j=0 C d,j t j )y d + B where B does not contain the term y d. Because d j=0 C d,jtj 0, there exists a non-zero number t Q such that d j=0 C d,j t j 0. Then d = tdeg(f (y, ty + z)) = deg(f (y, ty + z), y). It is not difficult to verify that (F (y, ty + z)) is still an irreducible polynomial and ȳ = P (x) Q(x), z = P (x)q(x) P (x)q (x) + tp (x)q(x) Q(x) 2 is a proper parametrization of F (y, ty + z) = 0. So d = deg(f (y, ty + z), y) = deg( z). By Lemma 3.4 and the same analysis as in the proof of Lemma 3.9, we get (1) If n > m, then d = deg(f (y, ty+z), y) = deg( z) = deg(ȳ 1 )+1 = deg(f (y, y 1 ), y)+1. (2) If n m, then d = deg(f (y, ty + z), y) = deg( z) = deg(ȳ 1 ) = deg(f (y, y 1 ), y). Furthermore, we have Theorem 4.3. Assume that F = 0 of the forms (4.1) and (4.2) has a nontrivial rational solution of form (4.3). Further assume that where C d,k C d,l C p,0 C p,0 0. Then F d(y, y 1 ) = C d, ky d ky k C d,k y d k y k 1, F d (y, y 1 ) = C d,d y d C d,ly d l y l 1, A 0 (y) = C p,0 y p + + C p,0 y p (4.4) (1) n > m iff d = p + 1. Moreover if n > m, then k = n m. (2) n < m iff d = p 1. Moreover if n < m, then l = m n.

13 Rational Solutions of ODEs 13 (3) n = m iff d = p and d = p. Proof. Case 1. Assume that n > m, then the Laurent serie expansion of the solution ȳ(x) = P (x) Q(x) at x = has the form: ȳ(x) = a n mx n m + + j=0 a jx j (see Section 4.2). Substituting ȳ(x) into F, for each monomial in F, the highest degree of C αi,β i ȳ(x) αi ȳ 1 (x) βi equals to (n m)(α i + β i ) β i. Let N = max{(n m)(α i + β i ) β i y αi y βi 1 appears in F }. Since F (ȳ(x)) = 0, there exist at least two monomials such that the highest degrees of them equal to N. Because p deg(f, y), by Theorem 4.2, p d 1. Suppose that p < d 1, then y d 1 y 1 must appear in F. However, N = (n m) d 1 > (n m)(α i + β i ) β i for all y αi y βi 1 except that y d 1 y 1, a contradiction. Hence p = d 1. From (4.4), we have (n m)( d 1) (n m)(α i + β i ) β i for all α i + β i d 1 and (n m) d k (n m) d β i for all α i + β i = d which imply (n m)( d 1) = (n m) d k. Hence k = n m. Now assume that d = p + 1. We will prove p = deg(f, y). Suppose that p deg(f, y) which implies that deg(f, y) > p. That is, in the form (4.1), there exists i 0 > 0 such that deg(a i0 (y)) > deg(a 0 (y)) = p. So we have d deg(a i0 (y)) + 1 > p + 1, a contradiction. Hence p = deg(f, y) which implies that d = deg(f, y) + 1. From Theorem 4.2, we have n > m. Case 2. Replace y by 1 z and y 1 by z1 z 2 in F and multiply F ( 1 z ) by z2d. By Lemma 4.1, we get an irreducible polynomial G(z). In G(z), the highest degree of the monomials is 2d d and the lowest degree of the monomials is 2d d. Corresponding to (4.4), we have Moreover, z = Q(x) P (x) G(z) = G 2d d (z) + G 2d d 1 (z) + + G 2d d(z) G 2d d (z) = C d,d z 2(d d) z d C d,lz 2d d l z l 1 G 2d d(z) = C d, kz 2d d kz k C d,k z 2d d k z k 1 Ā 0(z) = C p,0z 2d p + + C p,0z 2d p is a rational solution of G(z) = 0. By the first case, m > n iff 2d p = 2d d 1 and if m > n, then l = m n. In the other word, n < m iff d = p 1 and if m > n, then l = m n. Case 3. Assume n = m. By Theorem 4.2, d = deg(f, y) = max{deg(ai (y))}. Since d = max{deg(a i (y)) + i}, d = deg(a0 (y)) = p. As in the second case, we can obtain G(z), G 2d d (z), G 2d d(z) and Ā0(z). By the same reason, 2d d = 2d p which implies that d = p. The sufficiency is clear from the first case and the second case. In the special case m = 0, F will have the following particular type. Theorem 4.4. If F = 0 has a nontrivial rational solution of the form (4.3), then the nontrivial rational solution is polynomial iff F has the following form: F = ay n 1 + by n 1 + G(y, y 1 ) where n = deg(f, y 1 ), a, b Q are not zero, tdeg(g(y, y 1 )) n 1 and G does not contain the term y n 1. Proof. (= ) Since m = 0, from Theorem 4.3, k = n = deg(f, y 1 ). Hence F d(y) = C d,n y d n y n 1. From Lemma 4.1, d = n. By Theorem 4.3 again, p = n 1. So F has the above form. ( =) From Theorem 4.2, n > m. Since F d(y) = ay n 1, we have k = n m = n which implies m = 0.

14 14 Ruyong Feng and Xiao-Shan Gao 4.2. Computing the Laurent series soluton of F = 0 The first step of our algorithm to find the rational solutions to F = 0 is to find the its Laurent series solutions. We consider a Laurent series of the following form y(x) = i=k a i x i (4.5) where k is an integer and a i are in Q. Assuming that F = 0 has a nontrivial rational solution of the form (4.3) the above theorem provides a method to compute the Laurent series expansion of some solution of F = 0 at x =. Theorem 4.5. Use the notations in (4.1), (4.2), and (4.4). Assume that F = 0 has a nontrivial rational solution of form (4.3). Substituting (4.5) into F, we obtain a new Laurent series F (y(x)) = where L i are polynomials in a j. We have i=m L i x i (1) If d = p + 1, then in (4.5) let k = k, a k = C p,0 k k C d,k. We have L k( d 2)+i = C p,0 a d 2 k (k 1 i)a i + H i (a k 1,, a i+1 ) (4.6) where i = k 1, k 2,, and H i Q[a k 1,,ai+1 ]. Moreover, H k 1 (a k ) = 0. (2) If d = p and d = p 1, then in (4.5) let k = l, a l = ( l)l C d,l C p,0. We have L ld+i = C p,0 a d l (l i)a i + K i (a l 1,, a i+1 ) (4.7) where i = l 1, l 2,, and K i Q[a l 1,,ai+1 ]. Moreover, K l 1 (a l ) = 0. (3) If d = p and d = p, we can find a c Q such that F (y + c) satisfies the condition in case 2. Proof. We will prove the first case. The second case can be proved in the same way. Note that p = deg(f, y). First, we introduce some notations. Let A = C d,k y(x) d k y 1 (x) k + C p,0 y(x) d 1, B = F (y(x)) A. We will define a weight w : Z[x, a i ] Z which satisfies w(st) = w(s) + w(t) and w(x) = 1, w(a i ) = k i. Then y(x) is an isobaric polynomial with the weight k and y 1 (x) is an isobaric polynomial with the weight k 1. Hence A is an isobaric polynomial with the weight k( d 1) and the highest weight in B is less than k( d 1). Then the weight of L k( d 2)+i is less than or equal to k i. Therefore, a j can not appear in L k( d 2)+i for j < i and a i can only appear linearly in the coefficient of x k( d 2)+i in A. By the computation process, in the coefficient of x k( d 2)+i in A, the terms containing a i are C p,0 a d 2 k (k 1 i)a i. Since a i can not appear in the coefficient of x k( d 2)+i in B, L k( d 2)+i has the form (4.6). In the following, we prove H k 1 (a k ) = 0.

15 Rational Solutions of ODEs 15 From Theorem 4.3, we have that n > m. Then the Laurent series expansion of ȳ(x) at x = will have the form ȳ(x) = ā n m x n m + + ā j x j Moreover, by the computation, we have ā n m = C p,0 = a k k C k. Now substituting ȳ(x) d,k to F, we have L i = 0 for all i k( d 1) where L i are obtained by replacing a i with ā i in L i in (4.6). In particular, L k( d 1) 1 = H k 1 (a k ) = 0. We now prove the third case. From Theorem 4.3, we have n = m. Then the Laurent series expansion of the solution ȳ(x) at x = will have the form j=0 ȳ(x) = ā 0 + ā 1 1 x + ā 2( 1 x )2 + Then (ā 0, 0) will be a zero of F = 0 if we regard F as an algebraic polynomial. Hence A 0 (ā 0 ) = 0. From Theorem 3.14, there is a rational solution of F = 0 whose coefficients are in Q, assume that this solution is ȳ(x). Then ā 0 must be a rational number. Hence ā 0 is a rational root of A 0 (y) = 0. It is easy to know that F (y +ā 0 ) is still irreducible and ȳ(x) ā 0 is one of the solutions of F (y + ā 0 ) = 0. Because ȳ( ) ā 0 = 0, the degree of the numerator of ȳ(x) ā 0 is less than that of its denominator. Hence F (y + ā 0 ) should satisfy the condition of case 2. From Lemma 4.1, Theorem 4.2 and Theorem 4.5, we have an algorithm to compute the first m terms of a Laurent series solutions of F = 0 in some special case. Algorithm 4.6. Input: F and a positive integer m. Output: The first m terms of the Laurent series solution of F = 0 of form (4.5) or a message: F = 0 has no nontrivial rational solution. 1 Rewrite F as the form (4.1), (4.2) and (4.4). 2 Let d i = deg(a i (y)). For all i = 0,, d, if d i 2(d i), then go to next step. Otherwise by Lemma 4.1, F = 0 has no nontrivial rational solution; the algorithm terminates. 3 If d = p + 1, let k := k, a k := C p,0 k k C d,k, and ȳ := a k x k. (a) C :=the coefficient of x k( d 1) 1 in F (ȳ). If C 0 then by (4.6), F = 0 has no rational solutions and the algorithm terminates, else a k 1 := 0. (b) i := k 2. while i k m + 1 do i C := the coefficient of x k( d 2)+i in F (ȳ). C ii a i := iii ȳ := ȳ + a i x i. iv i := i 1. (c) return(ȳ).. C p,0a d 2 (k 1 i) k 4 If d = p and d = p 1, let k := l, a l := ( l)l C d,l C p,0, and ȳ := a l x l.

16 16 Ruyong Feng and Xiao-Shan Gao (a) C :=the coefficient of x l(d+1) 1 in F (ȳ). If C 0 then by (4.7), F = 0 has no rational solutions and the algorithm terminates, else a l 1 := 0. (b) i := l 2. while i l m + 1 do i C := the coefficient of x l d+i in F (ȳ). C ii a i := iii ȳ := ȳ + a i x i. iv i := i 1. (c) return(ȳ). C p,0a d l (l+1+i). 5 If d = p and p = d, then let r 1,, r k be all solutions of A 0 (y) = 0 in Q. Let i:=1. while i k do (a) F := F (y + r i ). Rewrite F as the form (4.1), (4.2) and (4.4). (b) If F satisfies the assumption of step 2 and step 4, then go to step 4. In the step (a) of the step 4, if C = 0, then run the steps (b) and (c) and return(ȳ + r i ), else go to the following step. (c) i := i + 1 If for all r i, we can not compute the first m terms Laurent series solution for F (y + r i ) = 0 in the step 4, then F = 0 has no nontrivial rational solution by Theorem 4.5 and the algorithm terminates. 6 In all other cases, F has no nontrivial rational solutions and the algorithm terminates. The complexity of computing ȳ d where ȳ is a polynomial in Q[x] with degree m is O(m 2 d 2 ). By Lemma 4.1, the total degree of F is at most 2d and the number of recurrence is at most 2md. The complexity of factorization of a polynomial with degree d in Q[x] is O(d 3 )(p. 411 (Gathen and Gerhard, 1999)). Hence Algorithm 4.6 is a polynomial time complexity algorithm. Here we only consider the number of multiplications (or divisions) in the algorithm. Remark 4.7. There is an algorithm based on the Newton Polygon method to compute Puiseux series solutions of differential equations (Cano, 1993; Duval, 1989; Grigor ev, 1991). In general, we need to solve the high degree algebraic equations to find the Puiseux series solutions by Newton Polygon method. Here, the differential equations which we consider are special. By the analysis of the structures of these special differential equations, we can determine the first term of one of its Laurent series solutions from the degrees and the coefficients of the original equation. Then we need only to use rational operations in Q to find all of the coefficients of its Laurent series solutions Padé Approximants The Padé approximants are a particular type of rational fraction approximation to the value of a function. It constructs the rational fraction from the Taylor series expansion of the original function. Its definition is given below (George and Baker, 1975). Definition 4.8. For the formal power series A(x) = 0 a jx j and two non-negative

17 Rational Solutions of ODEs 17 integers L and M, the (L, M) Padé approximant to A(x) is the rational fraction such that [L \ M] = P L(x) Q M (x) A(x) P L(x) Q M (x) = O(xL+M+1 ) where P L (x) is a polynomial with degree not greater than L and Q M (x) is a polynomial with degree not greater than M. Moreover, P L (x) and Q M (x) are relatively prime and Q M (0) = 1. Let P L (x) = L 0 p ix i and Q M (x) = M 0 q ix i. We can compute P L (x) and Q M (x) with the following linear equations in p i and q i : a 0 = p 0 a 1 + a 0 q 1 = p 1 a L + a L 1 q a 0 q L = p L a L+1 + a L q a L M+1 q M = 0 a L+M + a L+M 1 q a L q M = 0 (4.8) where a n = 0 if n < 0 and q j = 0 if j > M. For the Padé approximation, we have the following theorems (George and Baker, 1975). Theorem 4.9. (Frobenius and Padé) When it exists, the Padé approximant [L \ M] to any formal power series A(x) is unique. Theorem (Padé) The function f(x) is a rational function of the following form f(x) = p lx l + p l 1 x l p 0 q m x m + q m 1 x m iff the Padé approximants are given by [L \ M] = f(x) for all L l and M m A polynomial-time algorithm Let f(x) be a rational function. Rewrite f(x) as the form: f(x) = x k P (x) Q(x) where k Z and P (0) 0, Q(0) 0. Suppose that a 0 + a 1 x + is the Taylor series expansion of P (x) Q(x) at x = 0. Then by the uniqueness of the Laurent series expansion, xk (a 0 +a 1 x+ ) is the Laurent series expansion of f(x) at x = 0. So for a rational function g(x), if i=k a ix i is its Laurent series expansion at x = 0, then i=k a ix i k will be the Taylor series expansion of x k g(x) at x = 0. Then we can find x k g(x) by constructing Padé approxmiants to i=k a ix i k. This means that to find a rational function, we need only to know its Laurent series expansion at x = 0. Since the Laurent series expansion of g(x) at x = is equivalent to the Laurent series expansion of g( 1 t ) at t = 0, in order to find g(x), we need only to find the Laurent series expansion of g(x) at x =. In Section 4.2, we presented a method to compute a Laurent series solution of F = 0 at x =. Now, we are ready to give the main algorithm.

18 18 Ruyong Feng and Xiao-Shan Gao Algorithm Input is F. Output is a rational general solution of F = 0 if it exists. Also, if we find such a solution, it is of the following form ŷ = a n(x + c) n + a n 1 (x + c) n a 0 b m (x + c) m + b m 1 (x + c) m b 0 where a i, b j are in Q and c is an arbitrary constant. 1 d := deg(f, y 1 ). Compute the first 2d + 1 terms of Laurent series solution of F = 0 by Algorithm 4.6. If it returns a series ȳ(x), then go to next step, else the algorithm terminates. 2 Select an integer k such that z(t) := t k ȳ( 1 t ) is a polynomial and the first term of z(t) is a non-zero constant. 3 a i := the coefficient of t i in z(t) for i = 0,, 2d. In (4.8), let L = M = d. Then we can find q i by solving the following linear equations (note that we have q 0 = 1): a 1 a 2... a d a 2 a 3... a d a d a d+1... a 2d 1 q d q d 1. q 1 = a d+1 a d+2 If the above linear equation has no solutions, then the algorithm terminates. Otherwise, if the matrix A is singular, from Theorem 4.9, we need only to select one of the solutions of the above linear equations. 4 p i := a 0 q i + a 1 q i a i q 0 for i = 0 d and r(t) := p dt d + p d 1 t d p 0 q d t d + q d 1 t d ȳ(x) := x k r( 1 x ). Substituting ȳ(x) to F, if F (ȳ) = 0 then return ŷ = (x+c)k r( 1 x+c ). Otherwise, F = 0 has no rational general solution. By Lemma 2.6 and Theorem 2.7, we know that if F = 0 has a nontrivial rational solution, then every nontrivial formal Laurent series solutions of F = 0 must be the Laurent series of the rational solution. From Algorithm 4.6, we know that the Laurent series is nontrivial. By Theorem 3.10 and the discussion at the beginning of this subsection, the above algorithm is correct. Now we give an example to show how Algorithm 4.11 works. Example Consider differential equation: 1 Rewrite F as the form (4.2) F = y y (27y 2 + 4)y y 4 + 4y 2 = 0.. a 2d F = 27y 4 + y y 2 y + 4y y 2 + 4y 1 and A 0 (y) = 27y 4 + 4y 2 2 Use the notation in Theorem 4.5. Since d = p = 4 and d = 1 = p 1, F is in case 2 of Theorem d := deg(f, y 1 ) = 3. By Algorithm 4.6, compute the first 7 terms of the Laurent series solution ȳ(x) of F = 0: ȳ(x) = 1 x + 1 x 3

19 Rational Solutions of ODEs 19 deg(f, y 1 ) tdeg(f ) Average Time Rational Solutions no no no no no Table 1. Timings for randomly generated first order autonomous ODEs 4 k := 1 and z(t) := ȳ( 1 t ) t = 1 + t 2 5 a 0 := 1, : a 2 := 1 and a i := 0 for i = 1,, 7 and i 2. Solve the linear equations: ( ) ( q3 q 2 q 1 ) = ( Select one of the solutions as (q 3, q 2, q 1 ) := (0, 0, 0). 6 p 0 := 1, p 1 := 0, p 2 := 1, p 3 := 0 and r(t) = t ȳ(x) := r( 1 x ) x = x2 +1 x. Substituting ȳ(x) to F, F (ȳ(x)) 0. Hence ŷ(x) = (x+c) (x+c) 3 is a rational general solution. The complexity of the algorithm is clearly polynomial in terms of the number of multiplications (or divisions) in Q, since all the involved operations have polynomial complexity. In Step 1, the complexity is polynomial. In Step 3, we need to solve a linear equation system with rational coefficients and in 2n variables, which has polynomial complexity. In Step 5, we need to substitute y and y 1 in F (y, y 1 ) by two rational functions. From Theorem 3.10, deg(f, y) 2n. Then this step also has polynomial complexity. It is easy to check that all other steps are of polynomial complexity too Experimental results We implement Algorithm 4.11 in Maple. Two sets of experiments are performed. In the first experiment, we randomly generate two hundred first order autonomous ODEs of deg(f, y 1 ) = d and tdeg(f ) 2d for each d and compute the rational solutions for these equations. The result of the experiment shows that almost all of first order autonomous ODEs have no rational general solutions. The average running time is given in Table 1. Times are collected on a PC with a 2.66G CPU and 256M memory and are given in seconds. no in the last low of the table means there is no ODEs which have a rational general solution among two hundred ODEs which we compute. From Table 1, we can see that the program gives a negative answer immediately. The reason is that we can decide that the ODEs do not have rational general solutions using the necessary conditions presented in Section 4 without computation. In the second experiment, we generate first order autonomous ODEs having rational solutions based on Theorem 3.8 and then compute their rational solutions. Table 2 shows the computing times of the program for six examples. All of these examples are the case they have rational general solution. Times are collected on a PC with a 2.66G CPU and 256M memory and are given in seconds. We can see that the algorithm is generally very fast. In the table, deg means deg(f i, y 1 ), tdeg meas tdeg(f i ), term means the number of terms in F i, sol means whether F i has rational general solutions or not. The differential equations F i = 0 are given in the appendix of this paper. From Table 1, we can see that the new algorithm gives immediate negative answer to )

20 20 Ruyong Feng and Xiao-Shan Gao deg tdeg term time(s) sol deg tdeg term time(s) sol F yes F yes F yes F yes F yes F yes Table 2. Timings for solving first order autonomous ODEs almost all randomly generated ODEs. This is due to the structural properties in Theorem 4.5. From Table 2, we can see that the average running time of the new algorithm for first order autonomous ODEs with rational solutions is about thirty times faster than the old version in (Feng and Gao, 2004). 5. Conclusion In this paper, we give a necessary and sufficient condition for an ODE to have a rational general solution and a polynomial algorithm to compute the rational general solution of a first order autonomous ODE if it exists. As mentioned in Section 1, this work is motivated by the parametrization of differential algebraic varieties, which is still wide open. A problem of particular interests is to find conditions for a differential curve f(y, z) = 0 to have rational differential parameterizations. Developing effective algorithms to compute rational solutions for ODEs of the form y = R(x, y) is also very interesting. We may further ask whether we can define a differential genus for a differential curve similar to the genus of algebraic curves. References Abramov, S. and Kvashenko, K. (1991). Fast algorithm to search for the rational solutions of linear differential equations with polynomial coefficients, Proc. ISSAC1991, , ACM Press. Abhyankar, S.S. and Bajaj, C. (1988). Automatic parameterization of rational curves and surfaces, III: algebraic plane curves, Comp. Aided Geo. Design, 5, Bronstein, M. (1992). Linear ordinary differential equations: breaking through the order 2 barrier, Proc. ISSAC1992, 42-48, ACM Press. Bronstein, M. and Lafaille, S. (2002). Solutions of linear ordinary differential equations in terms of special functions, Proc. ISSAC2002, ACM Press. Bronstein, M. (1990). Integration of Elementary Functions, J. Symobolc Computation, 9, Cerveau, D. and Lins Neto, A. (1991). Holomorphic foliations in CP (2) having an invariant algebraic curve, Ann. Inst. Fourier, 41(4), Carnicer, M.M. (1994). The Poincaré problem in the nondicritical case, Ann. Math., 140, Cano, J. (2003). An algorithm to find polynomial solutions of y = R(x, y), private communication. Cano, J. (1993) On the series defined by differential equations, with an extension of the Puiseux polygon construction to these equations, Analysis, 13, Cormier, O. (2001). On Liouvillian solutions of linear differential equations of order 4 and 5, Proc. ISSAC2001, , ACM Press. Davenport, J.H. (1981). On the Integration of Algebraic Functions, Lecture Notes in Computer Science, 102, Springer-Verge, New York. Duval, D. (1989). Rational Puiseux expansions, Comp. Math., 70, Eremenko A. (1998). Rational solutions of first-order differential equations, Ann. Acad. Sci. Fenn., Math., 23, Feng, R. and Gao, X.S. (2004). Rational general solutions of algebraic ordinary differential equations, Proc. ISSAC2004, , ACM Press, New York. Forsyth A.R. (1959). Theory of Differential Equations, Dover Publications Inc.. Gao, X.S. (2003). Implicitization for differential rational parametric equations, J. Symbolic Computation, 36(5),